package cn.db117.leetcode.solution9;

/**
 * 999. 车的可用捕获量
 * 在一个 8 x 8 的棋盘上，有一个白色车（rook）。也可能有空方块，白色的象（bishop）和黑色的卒（pawn）。它们分别以字符 “R”，“.”，“B” 和 “p” 给出。大写字符表示白棋，小写字符表示黑棋。
 * <p>
 * 车按国际象棋中的规则移动：它选择四个基本方向中的一个（北，东，西和南），然后朝那个方向移动，直到它选择停止、到达棋盘的边缘或移动到同一方格来捕获该方格上颜色相反的卒。另外，车不能与其他友方（白色）象进入同一个方格。
 * <p>
 * 返回车能够在一次移动中捕获到的卒的数量。
 *  
 * <p>
 * 示例 1：
 * <p>
 * <p>
 * <p>
 * 输入：[[".",".",".",".",".",".",".","."],[".",".",".","p",".",".",".","."],[".",".",".","R",".",".",".","p"],[".",".",".",".",".",".",".","."],[".",".",".",".",".",".",".","."],[".",".",".","p",".",".",".","."],[".",".",".",".",".",".",".","."],[".",".",".",".",".",".",".","."]]
 * 输出：3
 * 解释：
 * 在本例中，车能够捕获所有的卒。
 * 示例 2：
 * <p>
 * <p>
 * <p>
 * 输入：[[".",".",".",".",".",".",".","."],[".","p","p","p","p","p",".","."],[".","p","p","B","p","p",".","."],[".","p","B","R","B","p",".","."],[".","p","p","B","p","p",".","."],[".","p","p","p","p","p",".","."],[".",".",".",".",".",".",".","."],[".",".",".",".",".",".",".","."]]
 * 输出：0
 * 解释：
 * 象阻止了车捕获任何卒。
 * 示例 3：
 * <p>
 * <p>
 * <p>
 * 输入：[[".",".",".",".",".",".",".","."],[".",".",".","p",".",".",".","."],[".",".",".","p",".",".",".","."],["p","p",".","R",".","p","B","."],[".",".",".",".",".",".",".","."],[".",".",".","B",".",".",".","."],[".",".",".","p",".",".",".","."],[".",".",".",".",".",".",".","."]]
 * 输出：3
 * 解释：
 * 车可以捕获位置 b5，d6 和 f5 的卒。
 *  
 * <p>
 * 提示：
 * <p>
 * board.length == board[i].length == 8
 * board[i][j] 可以是 'R'，'.'，'B' 或 'p'
 * 只有一个格子上存在 board[i][j] == 'R'
 * <p>
 * 来源：力扣（LeetCode）
 * 链接：https://leetcode-cn.com/problems/available-captures-for-rook
 * 著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。
 *
 * @author db117
 * @date 2020/1/10/010 16:03
 */
public class Solution999 {

    public int numRookCaptures(char[][] board) {
        int row = -1, col = -1;
        int ans = 0;
        // 找到车
        over:
        for (int i = 0; i < 8; i++) {
            for (int j = 0; j < 8; j++) {
                if (board[i][j] == 'R') {
                    row = i;
                    col = j;
                    break over;
                }
            }
        }

        for (int i = row; i < 8; i++) {
            char c = board[i][col];
            if (c == 'B') {
                break;
            }
            if (c == 'p') {
                ans++;
                break;
            }
        }

        for (int i = row - 1; i >= 0; i--) {
            char c = board[i][col];
            if (c == 'B') {
                break;
            }
            if (c == 'p') {
                ans++;
                break;
            }
        }

        for (int i = col; i < 8; i++) {
            char c = board[row][i];
            if (c == 'B') {
                break;
            }
            if (c == 'p') {
                ans++;
                break;
            }
        }

        for (int i = col - 1; i >= 0; i--) {
            char c = board[row][i];
            if (c == 'B') {
                break;
            }
            if (c == 'p') {
                ans++;
                break;
            }
        }
        return ans;
    }
}
